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sesenic [268]
3 years ago
11

​A hose, of radius 0.018 m, is connected to a water faucet. The water pressure at the point where the hose connects to the fauce

t is 150,000 Pa. The other end of the hose is connected to an outlet of radius 0.011 m, where the water comes out. The total length of the hose and outlet is 15 m.
(a) An experiment shows that the water exists the outlet with a speed 5.2 m/s. What is the flow rate through the hose?
(b) What is the maximum height that the end of the outlet can he held above the faucet, and water still comes out?
(c) Suppose there is a small hole in the hose. How does that affect the speed at which the water comes out? Explain.

Explain each step!
Physics
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

The smaller the hole, the more force.

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True a proton carries a positive charge, a neutron carries a neutral charge and an electron carries a negative charge.
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The value of 1.0004 to the power 1 by 2 using Binomial approximation is​
IgorLugansk [536]

Given:

The given value is (1.0004)^{\frac{1}{2}}.

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

(1.0004)^{\frac{1}{2}}

It can be written as:

(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}

(1.0004)^{\frac{1}{2}}=1+\dfrac{1}{2}\times 0.0004      [\because (1+x)^n=1+nx]

(1.0004)^{\frac{1}{2}}=1+0.0002

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3 years ago
A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
Musya8 [376]

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=\dfrac{q}{t}

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}

n=162500000 t

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The number of electron passe per second will be 162500000.

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3 years ago
Why do we use microwaves to communicate berween earth and satellites
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7 0
3 years ago
Read 2 more answers
I NEED HELP PLEASE, THANKS! :)
mrs_skeptik [129]

Answer:

1. Largest force: C;  smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

F=K\dfrac{ q_{1}q_{2}}{r^{2}}

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write  

F=\dfrac{k}{r^{2}}

1. Net force on each particle

Let's

  • Call the distance between adjacent charges d.
  • Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

F_{A} : F_{B} : F_{C} : F_{D}  =  \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\

2. Ratio of largest force to smallest

\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}

7 0
3 years ago
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